# Does a rocket engine apply more power as the rocket's speed increases? [duplicate]

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To clarify my level of knowledge, I'm a high school student all the way through AP Physics C: Mechanics.

So, let's imagine that there is a rocket travelling through the vacuum of space (ignoring gravity, air resistance, and all that). It's constantly accelerating with a $500\,{\rm kN}$ engine. Suppose, at some time, that it's going at $1\,{\rm km}\,{\rm s}^{-1}$. Since ${\rm power} = {\rm force} \cdot {\rm velocity}$, the power being applied instantaneously by the engine should be $500\,000\,{\rm kW}$, right?

Suppose that the rocket's speed has doubled to $2\,{\rm km}\,{\rm s}^{-1}$. Isn't the instantaneous power from the engine now $1\,000\,000\,{\rm kW}$?

So what confuses me is this: if the previous two paragraphs are correct, then isn't the engine using energy at a higher rate simply by virtue of going faster? And doesn't that mean that it's using up fuel at higher rate? But how could it be using more fuel when it's still applying the same force?

## marked as duplicate by knzhou, stafusa, Kyle Kanos, Jon Custer, ChemomechanicsSep 18 '18 at 18:47

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## 2 Answers

So what confuses me is this: if the previous two paragraphs are correct, then isn't the engine using energy at a higher rate simply by virtue of going faster?

No, it's not. You are forgetting that rockets consume mass, and in the process, they produce a cloud of exhaust gas behind them. You need to look at the rocket plus exhaust cloud.

I'll assuming a rocket traveling in a straight line that is far from a gravitating body and that is moving at much less than the speed of light. This makes for a very simple form of Newtonian mechanics: The rocket plus exhaust cloud conserve linear momentum.

Suppose the exhaust leaves the rocket at a constant effective speed $u$ relative to the rocket and at a constant positive mass flow rate $\dot m$, and suppose the rocket's speed relative to some observer is $v(t)$, directed against the exhaust velocity. Conservation of momentum dictates that $m \dot v = \dot m u$.

What about energy? I'll leave the math up to you, but regardless of the observer, the kinetic energy of the rocket plus gas cloud system grows at a rate $\frac {dE_\text{tot}}{dt} = \frac 1 2 \dot m u^2$. How this is distributed between the rocket versus the gas cloud depends on the observer.

I'll once leave the math up to you again, but the time rate of change in kinetic energy of the rocket is $\frac 1 2 \dot m v (2u-v)$. In other words, the rocket itself loses energy once the rocket's speed is twice that of the relative exhaust speed (but directed against the exhaust).

• Well that's something I didn't know: what a fascinating result for when $v>2\,u$. – WetSavannaAnimal Mar 29 '17 at 6:39

As you said, P = Force.velocity

Hence, P = Force.Distance/Time

Since the Distance and the Force are always the same, the only way to make the rocket faster is decreasing time.

A two times faster rocket, implies:

2.Power = Force.2.Velocity

2.Power = Force.2.Distance/Time

wich is the same thing as:

2.Power = Force.Distance/(Time/2)

Remember that Power = Work/Time and that Work is the amount of force necessary to move an object from point A to point B. This means that Power is actually the amount of joules per second you need to move something through a given distance in a given time.

In this case, the Work stays the same, because the amount of Force being applied and the Distance being covered doesn't change. This happens because you're not increasing the Force being applied, instead, is like you were trying to apply the same Force, but two times more often.

In doing so, you're decreasing the Time necessary to cover this given Distance by half and, thus, doubling the amount of instantaneous power needed.

Wich means that, in order to make it happen, the engine would need two times the amount of energy.